# Hess' Law question help

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#1
Got given these questions as homework and I'm completely stumped, any ideas where to begin with question C??
(Text in red are my answers)

5. Octane burns in oxygen (air);
C8H18 (l) + 25/2 O2 (g)  8 CO2 (g) + 9 H2O (l)
A 1.0 g sample of octane is burnt in a bomb calorimeter containing 1.20 kg of water. The heat capacity of the calorimeter is 837 J K-1 and for water it is 4.18 J g-1 K-1. The temperature of the water rises from 25.0 to 33.2C.

(a) Briefly, explain what is meant by a bomb calorimeter. A diagram is not required.

A type of calorimeter designed to withstand high pressures, which uses electrical energy to ignite the fuel, which then heats up the surrounding air. This air is then used to heat up the surrounding water.

(b) Calculate the heat of combustion of octane in kJ g-1 and the heat of combustion in kJ mol-1. Use the same kind of calculation as in the experiment (part 2).

q= -(1200*4.18+87)*8.2 =-47994.6J
47994.6J/1000= 47.99kJ
47.99kJ g-1

(C) Carbon burns in oxygen;
C(s) + O2 (g)  CO2 (g)
The enthalpy (heat) of reaction is -393 kJ mol-1.
Water burns in oxygen;
H2(g) + ½ O2 (g)  H2O (l)
The enthalpy (heat) of reaction is -242 kJ mol-1.
Octane is formed from its elements;
8 C (s) + 9 H2(g)  C8H18 (l)
The enthalpy (heat) of reaction is +208 kJ mol-1.
Use the above three reactions to calculate the heat of combustion of octane using Hess’s law. Compare your answer with that obtained in part (b).
0
7 years ago
#2
Hesses law states that the energy is the same irrelevant of the route
Set out a hess cycle and just follow it through
1
#3
(Original post by oriane89)
Hesses law states that the energy is the same irrelevant of the route
Set out a hess cycle and just follow it through
I don't even know where to begin with that lmao!
0
#4
(Original post by oriane89)
Hesses law states that the energy is the same irrelevant of the route
Set out a hess cycle and just follow it through
Right, I tried something, how does this look??

25O2 + 8C + 9H2 -> C8H18 + 25O2

|..................|............ ............|
-393..........-242.................+208
|..................|............ ............|
8CO2 + 9H2O-------------------

So to get the answer, would I do...

(-393) + (-242) + 208? = -427

Or... Do I have to times the numbers by the moles of the product,
e.g.

(-393 x 8) + (-242 x 9) + 208 = -5114

????
0
7 years ago
#5
you need to take into account the moles of product you're producing; I'll draw up a cycle and explanation tonight and post it on here
0
7 years ago
#6
(Original post by lsaul95)
Right, I tried something, how does this look??

25O2 + 8C + 9H2 -> C8H18 + 25O2

|..................|............ ............|
-393..........-242.................+208
|..................|............ ............|
8CO2 + 9H2O-------------------

So to get the answer, would I do...

(-393) + (-242) + 208? = -427

Or... Do I have to times the numbers by the moles of the product,
e.g.

(-393 x 8) + (-242 x 9) + 208 = -5114

????
In the question it says you have to use the 3 equations given.
The enthalpy values given in the question are for the formation and you have to calculate the enthalpy change of combustion. So first write a balanced equation for combustion of octane: C8H18 +25/2O2-----> 8CO2 +9H2O
Then refer the attachment and follow the steps.
You will end up getting two routes.
Using the 2 routes you can calculate the enthalpy change of combustion for octane!
Hope that helps
0
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